Levi Civita symbol on Curl of Vector cross product

[Nikitin]

Homework Statement

Use the LC symbol to calculate the following: $$\nabla \times \frac{\vec{m} \times \hat{r}}{r^2}$$

Where $\vec{m}$ is just a vector, and $\hat{r}$ is the unit radial vector and $r$ is the length of the radial vector.

Homework Equations

On the Levi Civita symbol: http://folk.uio.no/patricg/teaching/a112/levi-civita/

The Attempt at a Solution

I've gotten to $\epsilon_{kij} \epsilon_{klm} m_l \frac{r^2 \partial_j r_m - r_m \partial_j r^2}{r^4}$, but I don't know how to take the derivatives..

 

[Fredrik]

I've gotten to $\epsilon_{kij} \epsilon_{klm} m_l \frac{r^2 \partial_j r_m - r_m \partial_j r^2}{r^4}$, but I don't know how to take the derivatives..

That result doesn't look right to me. How did you get it? Note that if $\hat r=\frac{\vec r}{r}$, then
$$\frac{\hat r}{r^2}=\frac{\vec r}{r^3}=\bigg(\frac{r_i}{(r_jr_j)^{\frac 3 2}}\bigg) \hat e_i.$$

 

[RUber]

On the link you provided, there is a proof for $A\times B \times C=B(A \cdot C) - C(A \cdot B)$ using the LC method.
If $\vec{m}$ is just a vector, does that mean constant, i.e. derivatives equal zero?
That should leave you with something like $\vec{m} ( \nabla \cdot \frac{\hat{r}}{r^2})$
You could also look at $\frac{\hat{r}}{r^2}=\frac{1}{x^2+y^2+z^2}\hat{r}$ so taking the partial derivatives should not be too much of a challenge.

 

[Nikitin]

Ruber, yes but here you have a curl along with a function ($r^2$) involved.

But shouldn't it be much easier to perform the derivation in spherical coordinates? This is what's creating the struggle.

That result doesn't look right to me. How did you get it? Note that if $\hat r=\frac{\vec r}{r}$, then
$$\frac{\hat r}{r^2}=\frac{\vec r}{r^3}=\bigg(\frac{r_i}{(r_jr_j)^{\frac 3 2}}\bigg) \hat e_i.$$

I just wrote up the definition, ordered the resulting equation and used the product rule to get started with the derivations. Note, my $r_m$'s are actually the $m$th coordinate of $\hat{r}$, of not $\vec{r}$.

So basically I started with this: $$\epsilon_{ijk} \partial_j \epsilon_{klm} m_l \frac{r_m}{r^2}$$, and used the product rule after ordering to arrive to the equation in the OP.

 

[Fredrik]

Note, my $r_m$'s are actually the $m$th coordinate of $\hat{r}$, of not $\vec{r}$.

But the derivatives are with respect to the mth coordinate of $\vec r$, so with your definition of $r_m$, you don't have $\partial_i x_j=\delta_{ij}$. Instead you have to do something like this (where $x_m$ is the mth component of $\vec r=r\hat r$):
$$\partial_i r_j =\frac{\partial}{\partial x_i} r_j =\frac{\partial}{\partial x_i}\frac{x_j}{\sqrt{x_kx_k}}.$$

 

[Nikitin]

But since $\hat{r}$ is just a one-component vector, shouldn't it be extremely straightforward to use spherical-coordinates $\partial_i$ on it? Why do you change to xyz coordinates?

 

[vela]

It doesn't seem so straightforward to me. In spherical coordinates, $\vec{r}$ depends on all three coordinates in a relatively complicated way whereas in cartesian coordinates you have simply $\partial_i x_j = \delta_{ij}$.

 

[Fredrik]

I never changed coordinates. I just used the definitions. The cross product is defined by $x\times y=(x\times y)_i e_i=(\varepsilon_{ijk}x_jy_k) e_i$, where the $e_i$ are the standard basis vectors, not the basis vectors associated with the spherical coordinate system. $\nabla\times f(x)$ is defined similarly, as $\big(\varepsilon_{ijk}\frac{\partial}{\partial x_j}f_k(x)\big) e_i$.

In this problem, we have
$$f(x)=m\times\frac{\hat r}{r^2}=m\times\frac{\vec r}{r^3},$$ and therefore
$$f_k(x)=\varepsilon_{kln}m_l\frac{x_n}{r^3}=\varepsilon_{kln}m_l\frac{x_n}{(x_px_p)^{\frac 3 2}}.$$ I don't know if there's a way to use both spherical coordinates and the Levi-Civita symbol. (Note that the problem requires you to use the latter). If you know such a way, you can certainly try it.

 

[Nikitin]

OK fine I'll just do it as you said. Thanks.